I am reading this article on maximal ideal spaces and there is this part that I don't quite understand very well, hope you guys can help me out.
"Let $M(A)$ denote the maximal ideal space of a commutative Banach algebra $A$. If $C$ is a Banach subalgebra of $A$ and $\lambda\in M(C)$, then the set $M_{\lambda}(A):=\{\xi\in M(A): \xi|_C=\lambda\}$ is called the fiber of $M(A)$ over $\lambda$. Hence for every Banach algebra $A\subset L^{\infty}(\mathbb{R})$ with $M(C(\dot{\mathbb{R}})\cap A)=\dot{\mathbb{R}}$ and every $\lambda\in\dot{\mathbb{R}},$ the fiber $M_{\lambda}(A)$ denotes the set of all characters (multiplicative linear functionals) of $A$ that annihilate the set $\{f\in C(\dot{\mathbb{R}})\cap A:f(\lambda)=0\}$. Where $\dot{\mathbb{R}}$ is the compactification by one point of the real line $\mathbb{R}$"
The parts that I don't understand is where it say that the maximal ideal space of an algebra can be equal to $\dot{\mathbb{R}}$, (quote...with $M(C(\dot{\mathbb{R}})\cap A)=\dot{\mathbb{R}}$), How is this possible?
And how do you prove that the fiber $M_{\lambda}(A)$ denotes the set of all characters (multiplicative linear functionals) of $A$ that annihilate the set $\{f\in C(\dot{\mathbb{R}})\cap A:f(\lambda)=0\}$?.